Write the set of real numbers in interval notation. {x | x ≥ −8}
1 answer:
The answer is: 
Which can be written as: [-8, infinity)
This is the interval from -8 to infinity. So -8 is our left most point on the number line and infinity being our right most. There is no boundary on the right side.
Note how the left side has a square bracket and the right side has a curved parenthesis. This isn't a typo. The square bracket tells the reader "include the endpoint -8 as part of the solution set". The parenthesis tells the reader "do NOT include the endpoint infinity as part of the solution set".
Rule: infinity and negative infinity is always paired with a parenthesis because these aren't numbers. It's impossible to reach infinity, therefore it's impossible to include it in the set of values. If you could include it, then that implies you ran out of numbers.
Which can be written as: [-8, infinity)
This is the interval from -8 to infinity. So -8 is our left most point on the number line and infinity being our right most. There is no boundary on the right side.
Note how the left side has a square bracket and the right side has a curved parenthesis. This isn't a typo. The square bracket tells the reader "include the endpoint -8 as part of the solution set". The parenthesis tells the reader "do NOT include the endpoint infinity as part of the solution set".
Rule: infinity and negative infinity is always paired with a parenthesis because these aren't numbers. It's impossible to reach infinity, therefore it's impossible to include it in the set of values. If you could include it, then that implies you ran out of numbers.
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Answer
f(t)=2cos(t)
Step-by-step explanation:
let's describe the situation first, a fly is starting 2 meters from the bulb and flies towards the bulb and when it is really close to the bulb, it flies 2 meters further away from the bulb this means that it reaches d = 0m (distance from the bulb) and then flies further 2 meters so d = -2m.
it this point the fly returns back and touches the bulb and flies away (ends it's oscillatory motion ). d = 0 again and story ends here.
here if we want to model this problem with time function , the cosine function seems the best fit with amplitude of 2, so the answer is f(t) = 2cos(t).
Now you can ask why cosine function? well if you look at the graph or the plot of the function it perfectly captures the physical situation going on here in this problem.
Domain is 0 to 2 because it is one complete cycle and the range is -2 to +2.